Suppose $\displaystyle f:\mathbb{R} \rightarrow \mathbb{R}$ is such that both $\displaystyle f\prime$ and $\displaystyle f\prime\prime$ exist for all $\displaystyle x \in \mathbb{R}$, so that Taylor's theorem tells us that, for each $\displaystyle a, h\in \mathbb{R}$ there is a $\displaystyle \theta$ such that $\displaystyle 0<\theta<1$ so that $\displaystyle f(a+h)=f(a)+hf\prime(a)+\frac{1}{2}h^2 f\prime\prime(a+\theta h)$. Suppose further that on [0,2] the inequalities $\displaystyle |f(x)|\leq 1$ and $\displaystyle |f\prime\prime(x)| \leq 1$ hold. Write down the Taylor expansions of f(0) and f(2) about the point $\displaystyle x \in [0,2]$, using the above form of Taylor's Theorem, with a remainder involving $\displaystyle f\prime\prime$. Hence prove that for all $\displaystyle x \in [0,2]$ we have $\displaystyle |f\prime(x)| \leq 2$.

I'm not sure about the first part, what I have is:

$\displaystyle f(0)=f(-h)+hf\prime(-h)+\frac{1}{2}h^2f\prime\prime(h(\theta-1))$ by setting a+h=0 and

$\displaystyle f(2)=f(2-h)+hf\prime(2-h)+\frac{1}{2}f\prime\prime(2+h(\theta-1))$ but I don't think this is correct. It says about the point x in [0,2] but I don't know how to apply this to the formula. I have no idea for the last part. Thanks!